Bessel Function: Real v Parameter for y=0

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SUMMARY

The discussion centers on the necessity for the parameter v in the Bessel function differential equation \(x^2y'' + xy' + (x^2 - v^2)y = 0\) to be real and non-negative when defining standard Bessel functions. While the differential equation itself imposes no constraints on v beyond being finite, standard Bessel functions require v to be real and non-negative to ensure valid solutions. The conversation highlights the broader context in which Bessel functions can indeed accommodate complex arguments and indices.

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why does the v parameter in the bessel function
x^2y``+xy`+(x^2-v^2)y=0 have to be real and nonegative?
 
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In the differential equation, there are no constraints on v (other than being finite) and solutions exist. If you want to call the solutions "standard Bessel functions" then the values of v are restricted. Generally, however, Bessel functions can have both complex arguments and indices.
 
thank you very much!
 

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